# Non-trivial solutions for cyclotomic polynomials

I am seeking algebraic expressions which solve a polynomial equation, in particular an arbitrary cyclotomic polynomial. Let us agree we are not talking about expressions such as $e^{2\pi/7}$. My problem is to distinguish between those expressions I am going to call "trivial" and "non-trivial".

Typically, algebraic expressions with radical signs admit of multiple choices for the value of the radical expression: two for a square root, three for a cube root, etc. I will define a "non-trivial" algebraic solution as an expression in radicals such that for any choice of values, the expression is always a solution of the equation. Example: for any quadratic equation, there are two choices for the square root, and both choices give correct solutions.

For the fifth cyclotomic polynomial, the solution (which I looked up) is given by the expression: $\frac{\sqrt5 - 1}4 + \frac{\sqrt{-5 - \sqrt5}}8$ This expression takes on four possible values, and each of them is in fact a true solution of the fifth cyclotomic polynomial. It might look like it takes on eight values because there are three choices of the square root, but two of them must be the same choice: you cannot take one value for the square root of five, and then later in the same expression switch to the other value.

On the other hand, we have an alternative solution for the same polynomial: $1^{1/5}$ I define this as a "trivial" solution because it takes on five possible values, but only four of them are true solutions of the fifth cyclotomic polynomial. This expression is, of course, a non-trivial solution of the non-irreducible polynomial $x^5 - 1 = 0$, but that's not the polynomial we're talking about.

I recently posted a similar question, which I believe was correctly answered by Paul Garret, but I am not at all sure that everyone who participated in the discussion was actually working on the same question. I hope this clarifies the situation.

My question, then, was and is: do all the cyclotomic polynomials have "non-trivial" solutions? If not, what is the smallest polynomial for which such a solution (a) is not known, or (b) can be shown not to exist? I cannot believe this question is nonsense, and I would be very surprised if it hasn't been answered in the literature.

EDIT I am extremely gratifying that you guys had some fun with my question. I have always loved this topic and I think it is the greatest miracle of all math that the six choices of the cubic formula (three cube roots and two square roots) can be made to logically and unambiguously collapse to exactly three values.

Because I have only one check mark to award for "accepted answer", from many deserving contributors, I award my coveted check-mark to Ben. Thanks again guys.

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This looks interesting... –  Ｊ. Ｍ. Dec 20 '11 at 4:35
Thank you Dylan for the LaTeX! –  Marty Green Dec 20 '11 at 5:09
I'm not sure I agree the question has been answered. Can someone produce a single expression for the six primitive 7th roots of unity (and no other superfluous numbers)? I tried maple, mathematica, and gap's radiroot without success (maple: no expression, mathematica: 12 numbers, gap: 12 numbers). –  Jack Schmidt Dec 21 '11 at 20:19
@JackSchmidt - if you choose the roots consistently the Cardano formula works every time! –  Ben Blum-Smith Dec 21 '11 at 21:02
@JackSchmidt - to clarify: you interpret $x^3+ax=b$ as $(u+v)^3 - 3uv(u+v)=u^3+v^3$; so $x=u+v$, $a=-3uv$, and $b=u^3+v^3$; then you have a quadratic in $u^3,v^3$ because you know their sum ($b$) and product ($-a^3/27$). So you get $u^3$ from the quadratic formula. (The other choice of square root gives you $v^3$.) Then you take a cube root for $u$ - but which cube root forces which cube root you pick for $v$ because you know $a=-3uv$. Then $x=u+v$. So changing the choice of square root just switches $u,v$ and the only legitimate choice is which of the 3 cube roots you take for $u$. –  Ben Blum-Smith Dec 21 '11 at 21:10

After more thought, I feel sure enough to claim in an answer that $1^{1/5}$ is not a "solution in radicals" to $x^5-1=0$ in the sense that is guaranteed to exist by the solvability of the equation's Galois group. I think the kind of solution that the solvability of the Galois group implies exists is exactly a "non-trivial" solution in your sense.

The "solution in radicals" that is guaranteed by the solvability of the Galois group of a polynomial $f$ is really a "root tower" over $\mathbb{Q}$: a tower of fields, beginning with $\mathbb{Q}$ and ending with a field containing $f$'s splitting field, in which each field is obtained from the last one by adjoining a $p$th root for some prime $p$.

$$\mathbb{Q}\subset\mathbb{Q}[r_1]\subset\dots\subset\mathbb{Q}[r_1,\dots,r_k]$$

where for each $r_j$ there is a prime $p_j$ such that $r_j^{p_j}$ was already in the previous field $\mathbb{Q}[r_1,\dots,r_{j-1}]$ but $r_j$ is not. The mechanism of the proof is that if the Galois group $G$ is solvable, then there is a composition series

$$G=G_0 \triangleright G_1\triangleright \dots \triangleright G_k=\{1\}$$

such that each factor group $G_{j-1}/G_j$ is cyclic of prime order $p_j$. By the fundamental theorem of Galois theory, there is thus a tower of fields

$$\mathbb{Q}=K_0 \subset K_1 \subset \dots \subset K_k=\mathrm{splittingfield}(f)$$

where $K_j$ has degree $p_j$ over $K_{j-1}$. By possibly adjoining some extra elements, it can be guaranteed that each field extension can be accomplished by adjoining a $p_j$th root to the previous field (in other words a root $r_j$ of the polynomial $x^{p_j}-a$ where $a$ is in the previous field), so that the tower has the above form (a "root tower"). Because extra elements may have been added, the final field may now be bigger than $f$'s splitting field. (I am sweeping some possibly pertinent details under the rug here: the extra elements you need to adjoin are precisely the $p_j$th roots of unity. The argument, when applied to cyclotomic polynomials, is saved from circularity by induction on the size of the primes.)

The key point is this: when $p_j$th roots are adjoined, the field extensions have degree $p_j$. This means that the polynomial $x^{p_j}-a$ of which $r_j$ is a root must be irreducible over $K_{j-1}$. If the polynomial were reducible, $r_j$ would be the root of a polynomial of degree lower than $p_j$ and $K_j=K_{j-1}[r_j]$ would have degree less than $p_j$ over $K_{j-1}$.

What I'm getting at is that the $p$th roots that are adjoined at each step in such a "solution by radicals" are all roots of polynomials $x^p-a$ that are irreducible over the previous field. In particular, $$\mathbb{Q} \subset \mathbb{Q}[1^{1/5}]$$ is not a solution of $x^5-1$ by radicals in this sense, because $x^5-1$ isn't irreducible over $\mathbb{Q}$. In fact, in this context as you note, $1^{1/5}$ doesn't even have an unambiguous meaning. It could be $1$, in which case this is not even a nontrivial field extension. On the other hand it could be any of the four primitive fifth roots of unity, and then the expression $\mathbb{Q}[1^{1/5}]$ would have a meaning determined up to isomorphism, but it still wouldn't be a "solution by radicals" because fifth roots of unity do not solve an irreducible polynomial of the form $x^p-a$.

Furthermore, the solutions to an equation furnished by a "solution by radicals" of this kind are always "non-trivial" in your sense. Since at each stage the adjoined root $r_j$ is a root of an irreducible polynomial over $K_{j-1}$, replacing it with any other root of this polynomial induces an automorphism of $K_j$ fixing $K_{j-1}$ pointwise. At the end of everything, the roots of $f$ (the original polynomial to be solved) are elements of the top field $\mathbb{Q}[r_1,\dots,r_k]$; thus they are rational expressions in $r_1,\dots,r_k$ with rational coefficients. Replacing any of the $r_j$'s with a different root of the same irreducible polynomial it was a root of thus induces an automorphism of the top field $\mathbb{Q}[r_1,\dots,r_k]$ that fixes $\mathbb{Q}$ pointwise. Thus any expression in $r_1,\dots,r_k$ that solves $f$ will still solve $f$ after this replacement.

To make the connection to the question explicit, this means that the cyclotomic polynomials, because they have solvable Galois groups, do all have legitimate / "non-trivial" solutions in radicals, and this result dates back to the days of Gauss and Galois.

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I think the root tower requires roots of unity be adjoined first (so you cannot start with Q). For the 7th root of unity, one adjoins a 2nd and 3rd root of unity, and then one gets a non-trivial expression for the six primitive roots over Q[ζ3], but ζ3 itself has 2 values over Q, so I get some superfluous solutions. –  Jack Schmidt Dec 21 '11 at 20:22
Ah, got it. (a) the values of ζ3 are still switched by field automorphisms so your argument still works, and (b) I used inconsistent values of ζ3 in my explicit calculation, but when I kept them explicit the 12 possible values collapse to only 6 distinct values. –  Jack Schmidt Dec 21 '11 at 20:41
JackSchmidt's correction is exactly the point: there is an apparent superfluity of choices, but when they are made compatibly, the right number of solutions appears. Yes, the Lagrange resolvent algorithm presumes that pth roots of unity are in the groundfield to take a pth root. Thus, $1^{1/5}$ would accomplish nothing, because a 5th root would only be considered when 5th roots of_unity (which are, as Ben B-S notes, of lower degree) are already in the ground field. The weird feature of using seemingly irrelevant roots of unity is inescapable. –  paul garrett Dec 21 '11 at 20:51
The fact that we usually begin with $p$th roots of unity already in the ground field is what I was alluding to with "sweeping some possibly pertinent details under the rug." Paul alludes to something I missed when first learning this stuff: induction on the prime $p$, so we can guarantee we have adjoined $p$th roots of unity before we need to take a $p$th root, is an essential part of the argument that solvability of the Galois group over $Q$ implies solvability in radicals over $Q$. Without this, we just end up proving solvability in radicals over $Q$ adjoin all roots of unity. –  Ben Blum-Smith Dec 21 '11 at 21:00
In view of the comments, I would suggest editing the answer. It simply isn't true that the tower of fields ends with the splitting field of $f$; you may have to go to a field bigger than that to get the expression for roots of $f$ in radicals. The roots of a cyclic cubic, for example, can't be in ${\bf Q}(\root3\of b)$ for some rational $b$; $b$ must be in a quadratic extension of $\bf Q$. –  Gerry Myerson Dec 22 '11 at 5:00

The Galois group over the rationals of the field obtained by adjoining a primitive $n$th root of unity is abelian, a fortiori solvable. It follows that any primitive $n$th root of unity can be expressed in radicals, and the expression will be "non-trivial" in the sense in which you are using the word. Finding the radical expression can be a nuisance even for medium-sized values of $n$, and I do not know a reference where you can go to find such expressions.

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On that note, Mathematica has a hidden function called TrigToRadicals[] (available from the Developer package) that expands trigonometric values to radicals. The results are, as Gerry says, complicated. Try TrigToRadicals[Cos[Pi/23]], for instance. –  Ｊ. Ｍ. Dec 20 '11 at 4:37
I can't get over the logical problem in Gerry's answer, which is the same problem I had with Qiaochu's answer to my previous question: yes, the Galois group is rational, hence the polynomial is solvable, hence the root is expressible in radicals: but my problem is that "the nth root of 1" is ALSO an expression in radicals so how can you simply conclude from this line of reasoning that (hence) "the expression will be "non-trivial" in the sense that (I) am using the word"? –  Marty Green Dec 20 '11 at 14:13
J.M.'s comment effectively answers my question, because if Mathematica has a program that automatically generates the solution, then there must be some finite algorithm which reliably does the job. Therefore there is no sense in talking about the "lowest order cyclotomic which has not yet been solved in a non-trivial way"; even if no one has written down the solution, there is a finite algorithm which will generate it. –  Marty Green Dec 20 '11 at 14:15
@Marty, if the Galois group of a particular irreducible polynomial is not solvable, there can be no "sporadic" solution in radicals. If the Galois group is solvable, then there is a solution in "genuine" radicals, that is, without calling $\root5\of1$ a radical. This last statement is not immediately obvious, and I regret that as I am on the road I can't give a good reference for this result. –  Gerry Myerson Dec 21 '11 at 17:50
@Ben, my apologies for posting my comment before reading your answer, which (subject to what's been pointed out in the comments to your answer) looks like it does the trick. –  Gerry Myerson Dec 22 '11 at 5:02